如何使用 php 和 codeigniter 隐藏检索到的数据值的重复部分
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【中文标题】如何使用 php 和 codeigniter 隐藏检索到的数据值的重复部分【英文标题】：How to hide repeating part of the retrieved data values using php and codeigniter 【发布时间】：2018-01-14 14:42:40 【问题描述】：由于这个 mysql 语句，我得到了下表
SELECT * FROM expenses, income
WHERE expenses.projectname=income.projectname
AND expenses.task=income.task
这些是我的项目表中的字段
这些是我的任务表的字段
在此表中，一个项目有许多任务。因此项目、客户、项目开始和结束日期列重复无意义。我怎样才能为所有任务只显示一次？如何在这里应用 php 隐藏逻辑？下图显示了我需要实现的目标。通过 MySQL 查询检索数据。但是我怎样才能隐藏不必要的重复值 这是 CodeIgniter 视图页面
<table class="table table-lg">
<thead >
<tr class="filters">
<th><input type="text" class="form-control" placeholder="Project" disabled></th>
<th><input type="text" class="form-control" placeholder="Employee" disabled></th>
<th><input type="text" class="form-control" placeholder="Task" disabled></th>
<th><input type="text" class="form-control" placeholder="Expense" disabled></th>
<th><input type="text" class="form-control" placeholder="Amount" disabled></th>
<th><input type="text" class="form-control" placeholder="Paid/Not" disabled></th>
<th><input type="text" class="form-control" placeholder="Client" disabled></th>
<th><input type="text" class="form-control" placeholder="Cost" disabled></th>
<th><input type="text" class="form-control" placeholder="Income " disabled></th>
<th><input type="text" class="form-control" placeholder="Date" disabled></th>
</tr>
</thead>
<tbody>
<?php
if(isset($view_data1) && is_array($view_data1) && count($view_data1)): $i=1;
foreach ($view_data1 as $key => $data)
?>
<tr <?php if($i%2==0)echo 'class="even"';elseecho'class="odd"';?>>
<td><?php echo $data['projectname']; ?></td>
<td><?php echo $data['employee']; ?></td>
<td><?php echo $data['task']; ?></td>
<td><?php echo $data['ExpenseName']; ?></td>
<td><?php echo $data['ExpenseAmount']; ?></td>
<td><?php echo $data['pn']; ?></td>
<td><?php echo $data['cname']; ?></td>
<td><?php echo $data['taskcost']; ?></td>
<td><?php echo $data['amount']; ?></td>
<td><?php echo $data['datetimepicker_mask']; ?></td>
</tr>
<?php
$i++;
else:
?>
<tr>
<td colspan="7" align="center" >No Records Found..</td>
</tr>
<?php
endif;
?>
</tbody>
</table>
【问题讨论】：
您的图片链接已损坏 @JYoThI 不是链接坏了。请看我编辑的版本 也许 SQLGROUP BY 可以帮助您：dev.mysql.com/doc/refman/5.7/en/group-by-handling.html - 类似于 SELECT projectname, ANY_VALUE(task) FROM expenses, income WHERE expenses.projectname=income.projectname AND expenses.task=income.task GROUP BY projectname
@Dushee 感谢您的反馈...
@Dushee 你能记下两张表的所有列吗？我们很容易知道哪一列在两张表上。谢谢
【参考方案1】：
这是我的处理方式。我希望这和你遇到的问题一样:)
*重要，变量$dcost和$cd是我自己设置的变量，请根据你的情况调整
之前 后 代码前
foreach(your sql variable)
echo '<tr><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['destination'].'</td>
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['timein'].'</td>
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['timein'].'</td>
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['timein'].'</td>
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['timeout'].'</td>';
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" class="mday">'.$cd['mday'].'</td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" data-mrate="'.$cd['mrate'].'" class="mrate">'.rupiah($cd['mrate']).'</td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" class="tm"></td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" class="lday">'.$cd['lday'].'</td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" data-lrate="'.$cd['lrate'].'" class="lrate">'.rupiah($cd['lrate']).'</td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" class="tl"></td></tr>';
代码后
foreach(your sql variable)
echo '<tr>';
if($i - 1 == 1)
$i = 0;
echo ' <td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['destination'].'</td>
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['timein'].'</td>
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['timein'].'</td>
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['timein'].'</td>
<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;">'.$dcost['timeout'].'</td>';
else
echo ' <td colspan="5" style="border-right:1px solid #000;"> </td>';
echo '<td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" class="mday">'.$cd['mday'].'</td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" data-mrate="'.$cd['mrate'].'" class="mrate">'.rupiah($cd['mrate']).'</td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" class="tm"></td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" class="lday">'.$cd['lday'].'</td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" data-lrate="'.$cd['lrate'].'" class="lrate">'.rupiah($cd['lrate']).'</td><td style="border-right:1px solid #000;text-align:center;vertical-align:middle;" class="tl"></td></tr>';
【讨论】：
这没有提供问题的答案。您可以search for similar questions，或参考页面右侧的相关和链接问题找到答案。如果您有一个相关但不同的问题，ask a new question，并包含指向此问题的链接以帮助提供上下文。见：Ask questions, get answers, no distractions 对不起，我找到了我自己做的答案，请重新检查我已更改的 cmets【参考方案2】：使用in_array() 函数并将colspan 和rowspan 设置为td
在您的查看页面中使用以下代码，
<table class="table table-lg">
<thead >
<tr class="filters">
<th><input type="text" class="form-control" placeholder="Project" disabled></th>
<th><input type="text" class="form-control" placeholder="Employee" disabled></th>
<th><input type="text" class="form-control" placeholder="Task" disabled></th>
<th><input type="text" class="form-control" placeholder="Expense" disabled></th>
<th><input type="text" class="form-control" placeholder="Amount" disabled></th>
<th><input type="text" class="form-control" placeholder="Paid/Not" disabled></th>
<th><input type="text" class="form-control" placeholder="Client" disabled></th>
<th><input type="text" class="form-control" placeholder="Cost" disabled></th>
<th><input type="text" class="form-control" placeholder="Income " disabled></th>
<th><input type="text" class="form-control" placeholder="Date" disabled></th>
</tr>
</thead>
<tbody>
<?php
if(isset($view_data1) && is_array($view_data1) && count($view_data1)): $i=1;
foreach ($view_data1 as $key => $data)
$number_of_task[$data['projectname']][] = $data['task'];
foreach ($view_data1 as $key => $data)
$array = array();
?>
<tr <?php if($i%2==0)echo 'class="even"';elseecho'class="odd"';?>>
<?php if(!in_array($data['projectname'],$array))
$array[] = $data['projectname'];
?>
<td><?php echo $data['projectname']; ?></td>
<td><?php echo $data['employee']; ?></td>
<?php else ?>
<td colspan="2" rowspan="<?php echo count($number_of_task[$data['projectname']]); ?>"> </td>
<td><?php echo $data['task']; ?></td>
<td><?php echo $data['ExpenseName']; ?></td>
<td><?php echo $data['ExpenseAmount']; ?></td>
<td><?php echo $data['pn']; ?></td>
<td><?php echo $data['cname']; ?></td>
<td><?php echo $data['taskcost']; ?></td>
<td><?php echo $data['amount']; ?></td>
<td><?php echo $data['datetimepicker_mask']; ?></td>
</tr>
<?php
$i++;
else:
?>
<tr>
<td colspan="7" align="center" >No Records Found..</td>
</tr>
<?php
endif;
?>
</tbody>
</table>
【讨论】：
【参考方案3】：if($key == 0)
$project_name = $data['projectname'];
$first = true;
else
$first = false;
if($project_name != $data['projectname'])
$project_name = $data['projectname'];
$first = true;
<td><?php
if($first == true)
ehco $data['projectname'];
else
echo "";
</td>
在你的 foreach 中尝试这样，我没有厌倦它，但它可能会起作用，你可能会得到一些想法。在第一个循环中，它将检查 $key，首先它将为 0，如果 $key 不为 0，则 $first 变量将为真，然后它将检查 $project 名称到循环项目名称，如果它不匹配它将替换它.对您要删除或清空的客户端和其他列使用相同的内容。我希望这行得通，或者它能给你一些想法。
【讨论】：
【参考方案4】：首先，你应该重新排列 $viewdata1 数据结构。基本上，将主要项目数据字段和任务数据字段分开，然后以更好的方式将它们组合起来..
我的假设是： 1. [ task, ExpenseName, ExpenseAmount, pn, cname, taskcost, amount, datetimepicker_mask ] => 都是任务数据字段 2. [ projectname , employee ] => 都是项目数据字段
在您的 model（我们将其命名为 project_model）中，1. 有一个 ONLY 查询功能strong> [ projectname or employee or the projectID or all other project data fields ] ，我们将其命名为 allProject() 。确保它返回类似
[
0 => [
'id' => 101,
'projectName' => 'Desktop App' ,
other project data fields
] ,
1 => [
'id' => 102,
'projectName' => 'Mobile App' ,
other project data fields
] ,
2 => [
'id' => 103,
'projectName' => 'Andriod App' ,
other project data fields
] ,
etc...
]
2. 有一个基于 [ projectname 或 projectID ] 搜索/查询 [ task, ExpenseName, ExpenseAmount, pn, cname, taskcost, amount, datetimepicker_mask ] 的功能......我们来命名它task_query(projectID) ... 确保它返回类似
[
0 => [
'id' => 222,
'projectID' => 101
'task' => 'task 1',
'description' => 'description 1',
other task data fields
] ,
1 => [
'id' => 236,
'projectID' => 101
'task' => 'task 2',
'description' => 'description 2',
other task data fields
] ,
2 => [
'id' => 245,
'projectID' => 101
'task' => 'task 3',
'description' => 'description 3',
other task data fields
] ,
etc.
]
在控制器中，
$data['viewdata1'] = $this->project_model->allProject();
foreach($data['viewdata1'] as $key => $value)
$value['taskData'] = $this->project_model->task_query($value['id']);
$data['viewdata1'] 应该看起来像...
[
0 => [
'id' => 101,
'projectName' => 'Desktop App' ,
other project data fields,
'taskData' => [
0 => [
'id' => 222,
'projectID' => 101,
'task' => 'task 1',
'description' => 'description 1',
other task data fields
] ,
1 => [
'id' => 236,
'projectID' => 101,
'task' => 'task 2',
'description' => 'description 2',
other task data fields
] ,
2 => [
'id' => 245,
'projectID' => 101,
'task' => 'task 3',
'description' => 'description 3',
other task data fields
] ,
etc.
]
] ,
1 => [
'id' => 102,
'projectName' => 'Mobile App' ,
other project data fields,
'taskData' => [
0 => [
'id' => 123,
'projectID' => 102,
'task' => 'task 1b',
'description' => 'description 1b',
other task data fields
] ,
1 => [
'id' => 124,
'projectID' => 102,
'task' => 'task 2b',
'description' => 'description 2b',
other task data fields
] ,
2 => [
'id' => 125,
'projectID' => 102,
'task' => 'task 3b',
'description' => 'description 3b',
other task data fields
] ,
etc.
]
] ,
2 => [
'id' => 103,
'projectName' => 'Andriod App' ,
other project data fields,
'taskData' => [
0 => [
'id' => 567,
'projectID' => 103,
'task' => 'task 1c',
'description' => 'description 1c',
other task data fields
] ,
1 => [
'id' => 568,
'projectID' => 103,
'task' => 'task 2c',
'description' => 'description 2c',
other task data fields
] ,
2 => [
'id' => 569,
'projectID' => 103,
'task' => 'task 3c',
'description' => 'description 3c',
other task data fields
] ,
etc.
]
] ,
etc...
]
一切准备就绪
<table class="table table-lg">
<thead >
<tr class="filters">
<th><input type="text" class="form-control" placeholder="Project" disabled></th>
<th><input type="text" class="form-control" placeholder="Employee" disabled></th>
<th><input type="text" class="form-control" placeholder="Task" disabled></th>
<th><input type="text" class="form-control" placeholder="Expense" disabled></th>
<th><input type="text" class="form-control" placeholder="Amount" disabled></th>
<th><input type="text" class="form-control" placeholder="Paid/Not" disabled></th>
<th><input type="text" class="form-control" placeholder="Client" disabled></th>
<th><input type="text" class="form-control" placeholder="Cost" disabled></th>
<th><input type="text" class="form-control" placeholder="Income " disabled></th>
<th><input type="text" class="form-control" placeholder="Date" disabled></th>
</tr>
</thead>
<tbody>
<?php
if(isset($view_data1) && is_array($view_data1) && count($view_data1)): $i=1;
foreach ($view_data1 as $key1 => $data1)
$howManyTasks = count($data1['taskData']);
?>
<tr <?php if($i%2==0)echo 'class="even"';elseecho'class="odd"';?>>
<td rowspan="<?= $howManyTasks ?>"><?php echo $data1['projectname']; ?></td>
<td rowspan="<?= $howManyTasks ?>"><?php echo $data1['employee']; ?></td>
<?php foreach($data1['taskData'] as $key2 => $data2) ?>
<td><?php echo $data2['task']; ?></td>
<td><?php echo $data2['ExpenseName']; ?></td>
<td><?php echo $data2['ExpenseAmount']; ?></td>
<td><?php echo $data2['pn']; ?></td>
<td><?php echo $data2['cname']; ?></td>
<td><?php echo $data2['taskcost']; ?></td>
<td><?php echo $data2['amount']; ?></td>
<td><?php echo $data2['datetimepicker_mask']; ?></td>
</tr>
<?php if($key2 != ($howManyTasks-1) ) ?> <tr> <?php ?>
<?php ?>
<?php
$i++;
else:
?>
<tr>
<td colspan="7" align="center" >No Records Found..</td>
</tr>
<?php
endif;
?>
</tbody>
</table>
【讨论】：
【参考方案5】：将您的查询编辑为SELECT * FROM expenses, income WHERE expenses.projectname=income.projectname AND expenses.task=income.task Blockquote GROUP BY PROJECTID
【讨论】：
【参考方案6】：逻辑
我实现的隐藏逻辑，按以下顺序工作：
初始条件 主循环 条件检查 记住最后一个条件HTML 表和 PHP 如下所示：
<table border="1">
<thead>
<tr>
<th>Project</th>
<th>Client</th>
<th>Project Start</th>
<th>Project End</th>
<th>Task</th>
<th>Description</th>
<th>Commission</th>
<th>Task Start</th>
<th>Task End</th>
</tr>
</thead>
<tbody>
<!-- Initial Condition -->
<?php $the_project = "";?>
<!-- Main Loop -->
<?php foreach ($data as $x => $datum) ?>
<tr>
<!-- Condition Check -->
<!-- If not the same with the last project, it will show the columns. -->
<!-- That's why the Initial Condition is empty so that the 1st time this Condition Check run will always result true. -->
<?php if($datum['project'] != $the_project) ?>
<td><?php echo $datum['project'];?></td>
<td><?php echo $datum['client'];?></td>
<td><?php echo $datum['proj_start'];?></td>
<td><?php echo $datum['proj_end'];?></td>
<?php else ?>
<td colspan="4"> </td>
<?php ?>
<!-- /Condition Check -->
<td><?php echo $datum['task'];?></td>
<td><?php echo $datum['desc'];?></td>
<td><?php echo $datum['commission'];?></td>
<td><?php echo $datum['task_start'];?></td>
<td><?php echo $datum['task_end'];?></td>
</tr>
<!-- Remember Last Condition -->
<?php $the_project = $datum['project'];?>
<?php ?>
<!-- /Main Loop -->
</tbody>
</table>
您不必比较所有Project、Client、Start Date 和End Date，假设同一个项目将具有相同的Client、Start Date 和End Date。这就是为什么它只比较项目 ($the_project)。
数据
您可以将其用作示例数据，其作用类似于查询中的数据。
<?php
$data = array(
array(
"project" => "Mobile App",
"client" => "Client One",
"proj_start"=> "12-12-2017",
"proj_end" => "12-12-2019",
"task" => "task 1",
"desc" => "desc 1",
"commission"=> "1000",
"task_start"=> "01-01-2018",
"task_end" => "10-01-2018"
),
array(
"project" => "Mobile App",
"client" => "Client One",
"proj_start"=> "12-12-2017",
"proj_end" => "12-12-2019",
"task" => "task 2",
"desc" => "desc 2",
"commission"=> "1000",
"task_start"=> "01-02-2018",
"task_end" => "10-02-2018"
),
array(
"project" => "Mobile App",
"client" => "Client One",
"proj_start"=> "12-12-2017",
"proj_end" => "12-12-2019",
"task" => "task 3",
"desc" => "desc 3",
"commission"=> "1000",
"task_start"=> "01-04-2018",
"task_end" => "10-05-2018"
),
array(
"project" => "Web App",
"client" => "Client Two",
"proj_start"=> "12-12-2017",
"proj_end" => "12-12-2019",
"task" => "task 1",
"desc" => "desc 1",
"commission"=> "1000",
"task_start"=> "01-01-2018",
"task_end" => "10-01-2018"
),
array(
"project" => "Web App",
"client" => "Client Two",
"proj_start"=> "12-12-2017",
"proj_end" => "12-12-2019",
"task" => "task 2",
"desc" => "desc 2",
"commission"=> "1000",
"task_start"=> "01-02-2018",
"task_end" => "10-02-2018"
)
);
?>
提示 1
我认为 $i 不再需要，因为您可以在代码示例中从 $key 获得相同的结果。唯一的区别是$key从0开始。
提示 2
如果你还需要$i，你可以简化这段代码
<tr <?php if($i%2==0)echo 'class="even"';elseecho'class="odd"';?>>
进入
<tr class="<?php echo ($i%2==0) ? "even":"odd" ?>">
提示 3
我不知道<tr> 中的类偶数或奇数的目的，但如果您想为每个类设计不同的颜色，您可能需要检查Bootstrap 并将其设置为<table class="table table-striped">
【讨论】：
【参考方案7】：如果我理解正确，您可以尝试以下操作。
<?php
$arrData = [
[
"projectname" => "First Project",
"employee" => "First Client",
"projectstart" => "12-12-2017",
"projectend" => "12-12-2019",
"task" => "Task 1",
"description" => "Description 1",
"commission" => "1000",
"taststart" => "01-01-2018",
"taskend" => "10-01-2018"
],
[
"projectname" => "First Project",
"employee" => "First Client",
"projectstart" => "12-12-2017",
"projectend" => "12-12-2019",
"task" => "Task 2",
"description" => "Description 2",
"commission" => "1000",
"taststart" => "01-01-2018",
"taskend" => "10-01-2018"
],
[
"projectname" => "First Project",
"employee" => "First Client",
"projectstart" => "12-12-2017",
"projectend" => "12-12-2019",
"task" => "Task 3",
"description" => "Description 3",
"commission" => "1000",
"taststart" => "01-01-2018",
"taskend" => "10-01-2018"
],
[
"projectname" => "Second Project",
"employee" => "Second Client",
"projectstart" => "12-12-2017",
"projectend" => "12-12-2019",
"task" => "Task 4",
"description" => "Description 4",
"commission" => "1000",
"taststart" => "01-01-2018",
"taskend" => "10-01-2018"
],
[
"projectname" => "Second Project",
"employee" => "Second Client",
"projectstart" => "12-12-2017",
"projectend" => "12-12-2019",
"task" => "Task 5",
"description" => "Description 5",
"commission" => "1000",
"taststart" => "01-01-2018",
"taskend" => "10-01-2018"
],
[
"projectname" => "Second Project",
"employee" => "Second Client",
"projectstart" => "12-12-2017",
"projectend" => "12-12-2019",
"task" => "Task 6",
"description" => "Description 6",
"commission" => "1000",
"taststart" => "01-01-2018",
"taskend" => "10-01-2018"
],
];
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<!-- The above 3 meta tags *must* come first in the head; any other head content must come *after* these tags -->
<title>Bootstrap 101 Template</title>
<!-- Bootstrap -->
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" rel="stylesheet" integrity="sha384-1q8mTJOASx8j1Au+a5WDVnPi2lkFfwwEAa8hDDdjZlpLegxhjVME1fgjWPGmkzs7" crossorigin="anonymous">
<!-- HTML5 shim and Respond.js for IE8 support of HTML5 elements and media queries -->
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<![endif]-->
</head>
<body>
<div class="container">
<table class="table table-striped table-hover">
<thead>
<tr>
<th>Project</th>
<th>Client</th>
<th>Project start on</th>
<th>Project end on</th>
<th>task</th>
<th>description</th>
<th>commission</th>
<th>task start on</th>
<th>task end on</th>
</tr>
</thead>
<tbody>
<?php
$strSavedProjectName = false;
foreach($arrData AS $arrItem)
?>
<tr>
<?php
if (!$strSavedProjectName || $strSavedProjectName!= $arrItem['projectname']) :
?>
<td><?=$arrItem['projectname']; ?>
<td><?=$arrItem['employee']; ?>
<td><?=$arrItem['projectstart']; ?>
<td><?=$arrItem['projectend']; ?>
<?php
else :
?>
<td colspan="4"></td>
<?php
endif;
?>
<td><?=$arrItem['task']; ?>
<td><?=$arrItem['description']; ?>
<td><?=$arrItem['commission']; ?>
<td><?=$arrItem['taststart']; ?>
<td><?=$arrItem['taskend']; ?>
</tr>
<?php
$strSavedProjectName = $arrItem['projectname'];
?>
</tbody>
</table>
</div>
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<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
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</body>
</html>
这是一个基于您的数据的完整示例 - 因此您应该能够轻松地将其转换为您的视图。
【讨论】：
【参考方案8】：我没有测试这段代码，但想法是记住最后一个键（在本例中为项目名称和员工），然后将其与当前键进行比较。
您的示例代码中的字段与您的 Word 文档示例表不匹配，所以我只使用了您的代码示例。
<tbody>
<?php
if (isset($view_data1) && is_array($view_data1) && count($view_data1))
$i = 1;
$last_key = '';
foreach ($view_data1 as $key => $data)
$stripe = 'odd';
if ($i % 2 == 0)
$stripe = 'even';
echo "<tr class='$stripe'>";
$i++;
$current_key = $data['projectname'] . $data['employee'];
if ($current_key !== $last_key)
echo "<td>" . $data['projectname'] . "</td>";
echo "<td>" . $data['employee'] . "</td>";
else
echo "<td colspan='2'></td>";
$last_key = $current_key;
?>
<td><?php echo $data['task']; ?></td>
<td><?php echo $data['ExpenseName']; ?></td>
<td><?php echo $data['ExpenseAmount']; ?></td>
<td><?php echo $data['pn']; ?></td>
<td><?php echo $data['cname']; ?></td>
<td><?php echo $data['taskcost']; ?></td>
<td><?php echo $data['amount']; ?></td>
<td><?php echo $data['datetimepicker_mask']; ?></td>
</tr>
<?php
else
?>
<tr><td colspan="10" align="center" >No Records Found..</td></tr>
<?php
?>
</tbody>
</table>
【讨论】：
这可能是最简单的方法，或多或少地使用程序的其余部分。【参考方案9】：不要这样合并表格。在模型中使用连接查询并通过控制器在视图中调用它
public function getExpenses()
$this->db->select("addexpense.exp_date, addexpense.exp_amount, addexpense.exp_note, addexpense.exp_created, addcategory.category_name");
$this->db->from('addexpense');
$this->db->join('addcategory', 'addcategory.category_id = addexpense.category_id');
$query = $this->db->get();
return $query->result();
【讨论】：
【参考方案10】：不要同时选择两个表。加入他们的条件。
SELECT expenses.*,income.*,expenses.id as p_id FROM expenses
join income ON expenses.task=income.task AND expenses.projectname=income.projectname
更改每个部分的视图
<table class="table table-lg">
<thead >
<tr class="filters">
<th ><input type="text" class="form-control" placeholder="Project" disabled></th>
<th ><input type="text" class="form-control" placeholder="Employee" disabled></th>
<th ><input type="text" class="form-control" placeholder="Task" disabled></th>
<th ><input type="text" class="form-control" placeholder="Expense" disabled></th>
<th ><input type="text" class="form-control" placeholder="Amount" disabled></th>
<th ><input type="text" class="form-control" placeholder="Paid/Not" disabled></th>
<th ><input type="text" class="form-control" placeholder="Client" disabled></th>
<th ><input type="text" class="form-control" placeholder="Cost" disabled></th>
<th ><input type="text" class="form-control" placeholder="Income " disabled></th>
<th ><input type="text" class="form-control" placeholder="Date" disabled></th>
</tr>
</thead>
<tbody>
<?php
if(isset($view_data1) && is_array($view_data1) && count($view_data1)): $i=1;
$is_exists=array();
foreach ($view_data1 as $key => $data)
?>
<tr <?php if($i%2==0)echo 'class="even"';elseecho'class="odd"';?>>
<?php
if(!in_array($data['p_id'], $is_exists))
$is_exists[]=$data['p_id'];
?>
<td><?php echo $data['projectname']; ?></td>
<td><?php echo $data['employee']; ?></td>
<td><?php echo $data['task']; ?></td>
<td><?php echo $data['ExpenseName']; ?></td>
<?php
else
echo "<td rowspan='4'></td>";
?>
<td><?php echo $data['ExpenseAmount']; ?></td>
<td><?php echo $data['pn']; ?></td>
<td><?php echo $data['cname']; ?></td>
<td><?php echo $data['taskcost']; ?></td>
<td><?php echo $data['amount']; ?></td>
<td><?php echo $data['datetimepicker_mask']; ?></td>
</tr>
<?php
$i++;
else:
?>
<tr>
<td colspan="7" align="center" >No Records Found..</td>
</tr>
<?php
endif;
?>
</tbody>
</table>
【讨论】：
这是我的查看代码 Mr.Kundan。我没有使用 ms word 设计的表格的视图代码。我想更改上面的视图代码，以便获得我从 ms word 设计的表格。 非常感谢您的回答。但它没有任何改变 你是否也在 foreach.. 中加入 else 条件？ 让我们continue this discussion in chat。 Yes Mr.Kundan..else 如果没有记录会显示错误信息以上是关于如何使用 php 和 codeigniter 隐藏检索到的数据值的重复部分的主要内容，如果未能解决你的问题，请参考以下文章
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